Angular Momentum or Moment of Linear Momentum

Definition

The quantity in rotational mechanics analogous to linear momentum is angular momentum, or the moment of linear momentum. It stands to linear momentum as torque stands to force.

If $\vec p$ is the instantaneous linear momentum of a particle undertaking circular motion, its angular momentum at that instant is

$$\vec L = \vec r\times\vec p$$

where $\vec r$ is the position vector from the axis of rotation.

In magnitude it is the product of the linear momentum and its perpendicular distance from the axis of rotation:

$$L = p\,r\sin\theta = m v r\sin\theta$$

where $\theta$ is the smaller angle between the directions of $\vec p$ and $\vec r$.

• SI unit: kg m$^2$/s (equivalently J s)
• Dimensions: $[M^1L^2T^{-1}]$

Expression for Angular Momentum in Terms of Moment of Inertia

Diagram

The same figure as for rotational kinetic energy: a rigid body rotating with constant angular speed $\omega$ about an axis perpendicular to the plane of the paper, with particles $m_1, m_2, \dots, m_N$ at perpendicular distances $r_1, r_2, \dots, r_N$ from the axis, each with a tangential velocity.

Derivation

Consider a rigid object rotating with a constant angular speed $\omega$ about an axis perpendicular to the plane of the paper. For theoretical simplification let the object consist of $N$ particles of masses $m_1, m_2, \dots, m_N$ at respective perpendicular distances $r_1, r_2, \dots, r_N$ from the axis of rotation.

As the object rotates, all these particles perform U.C.M. with the same angular speed $\omega$ but with different linear speeds

$$v_1 = r_1\omega, \quad v_2 = r_2\omega, \quad \dots, \quad v_N = r_N\omega$$

The directions of the individual velocities are along the tangents to their respective tracks. The linear momentum of the first particle is of magnitude

$$p_1 = m_1v_1 = m_1r_1\omega$$

directed along $\vec v_1$. Its angular momentum is therefore of magnitude

$$L_1 = p_1r_1 = m_1r_1^2\omega$$

Similarly $L_2 = m_2r_2^2\omega$, $L_3 = m_3r_3^2\omega$, and so on up to $L_N = m_Nr_N^2\omega$.

For a rigid body with a fixed axis of rotation, all these angular momenta are directed along the axis of rotation, the direction being given by the right-hand thumb rule. As all of them have the same direction, their magnitudes can be added algebraically:

$$L = m_1r_1^2\omega + m_2r_2^2\omega + \dots + m_Nr_N^2\omega$$

$$L = \left(m_1r_1^2 + m_2r_2^2 + \dots + m_Nr_N^2\right)\omega = I\omega$$

where $I$ is the moment of inertia of the body about the given axis of rotation.

The expression $L = I\omega$ is analogous to $p = mv$, with the moment of inertia $I$ replacing the mass — which is the physical significance of $I$.

Kinetic Energy in Terms of Angular Momentum

The board sets this as a "show that" question, and the expected answer runs through the rotational kinetic energy derivation first.

As above, the translational kinetic energy of the first particle is $(K.E.)_1 = \dfrac{1}{2}m_1v_1^2 = \dfrac{1}{2}m_1r_1^2\omega^2$, and similarly for all the other particles. The rotational kinetic energy of the object is the sum of the individual translational kinetic energies:

$$\text{Rotational K.E.} = \dfrac{1}{2}m_1r_1^2\omega^2 + \dfrac{1}{2}m_2r_2^2\omega^2 + \dots + \dfrac{1}{2}m_Nr_N^2\omega^2$$

$$\therefore\ \text{Rotational K.E.} = \dfrac{1}{2}\left(m_1r_1^2 + m_2r_2^2 + \dots + m_Nr_N^2\right)\omega^2$$

But $I = \sum_{i=1}^{N} m_ir_i^2$, so

$$\text{Rotational K.E.} = \dfrac{1}{2}I\omega^2 \qquad \dots(i)$$

The angular momentum of a rotating body is

$$L = I\omega \qquad \dots(ii)$$

Writing (i) as $K.E. = \dfrac{1}{2}(I\omega)\omega$ and substituting $I\omega$ from (ii),

$$K.E. = \dfrac{1}{2}L\omega$$

Equivalently, substituting $\omega = \dfrac{L}{I}$ gives $K.E. = \dfrac{L^2}{2I}$.

Standard Results

• $\vec L = \vec r\times\vec p$; $\quad L = pr\sin\theta = mvr\sin\theta$
• $L = I\omega$, analogous to $p = mv$
• $K.E. = \dfrac{1}{2}I\omega^2 = \dfrac{1}{2}L\omega = \dfrac{L^2}{2I}$
• SI unit kg m$^2$/s; dimensions $[M^1L^2T^{-1}]$
• $\vec\tau = \dfrac{d\vec L}{dt}$ — the rotational analogue of $\vec f = \dfrac{d\vec p}{dt}$